Đặt a/2=b/3=c/4=k
=>a=2k; b=3k; c=4k
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)
\(\Leftrightarrow k^2=16\)
Trường hợp 1: k=4
=>a=8; b=12; c=16
Trường hợp 2: k=-4
=>a=-8; b=-12; c=-16
REFER
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=64\\b^2=144\\c^2=256\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\pm8\\b=\pm\\c=\pm16\end{matrix}\right.12}\)
Vậy (a; b; c)\(\in\){(8; 12; 16)}; {(-8; -12; -16)}
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\left\{{}\begin{matrix}b=\dfrac{3}{2}a\\c=2a\end{matrix}\right.\).
Ta có: \(a^2+3b^2-2c^2=a^2+3.\left(\dfrac{3}{2}a\right)^2-2.\left(2a\right)^2=-\dfrac{1}{4}a^2=-16\) \(\Rightarrow\) a=\(\pm\)8 \(\Rightarrow\) b=\(\pm\)12, c=\(\pm\)16.
làm lại cái lỗi:
\(\left\{{}\begin{matrix}a^2=64\\b^2=144\\c^2=256\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}a=\pm8\\b=\pm12\\c=\pm16\end{matrix}\right.\)
hy vọng bạn có thể bỏ qua cái lỗi phép tính này 
